Asked by BARBIE LEE
imageshack(.)(us)/photo/my-images/834/j6wv.jpg/
imageshack(.)(us)/photo/my-images/17/6n4x.jpg/
please PLEASE IM BEGGING YOU MY TEACHER WILL BE GIVING ME 79 IF I DIDN'T GET THIS .. OR ELSE LOWER THAN THAT. PLEASE
imageshack(.)(us)/photo/my-images/17/6n4x.jpg/
please PLEASE IM BEGGING YOU MY TEACHER WILL BE GIVING ME 79 IF I DIDN'T GET THIS .. OR ELSE LOWER THAN THAT. PLEASE
Answers
Answered by
BARBIE LEE
just delete the parethesis please
Answered by
Reiny
The key to this question is to realize that all the right -angled triangles are similar, and have the
ratio of 3:4:5 in the length of their sides.
Mark the two angles in triangle BCD with o and x
( you should be able to see that since XD | BC , so you can mark XDC as o and DCX as x
continue through the other triangles...)
Triangle XDC is similar to triangle DCB
XD:XC:DC = DC:DB:CB = 4:3:5
4p:3p:4
4p/4 = 4/5
4p = 16/5 ----> XD = 16/5
3p/4 = 3/5
3p=12/5 ----> XC = 12/5
and of course ---> CD = 4
now you have all the sides of triangle XDC
Did you notice that
3:4:5 = (5/4)(12/5 : 16/5 : 4) ??
continue in this fashion until you have all sides needed.
ratio of 3:4:5 in the length of their sides.
Mark the two angles in triangle BCD with o and x
( you should be able to see that since XD | BC , so you can mark XDC as o and DCX as x
continue through the other triangles...)
Triangle XDC is similar to triangle DCB
XD:XC:DC = DC:DB:CB = 4:3:5
4p:3p:4
4p/4 = 4/5
4p = 16/5 ----> XD = 16/5
3p/4 = 3/5
3p=12/5 ----> XC = 12/5
and of course ---> CD = 4
now you have all the sides of triangle XDC
Did you notice that
3:4:5 = (5/4)(12/5 : 16/5 : 4) ??
continue in this fashion until you have all sides needed.
Answered by
Anonymous
yes sir ive got that already. i don't know what to do next. please help
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